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Readme [11.4K]
2 years ago
14

A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th

at she traveled 0.1 mi while accelerating. Determine (a) the acceleration of the car, (b) the time required to reach 65 mi/h.
Engineering
1 answer:
Helga [31]2 years ago
5 0

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

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Answer :

The mass of perchloric acid is, 26.5 grams.

The mass of water in the same solution is, 11.1 grams

Explanation :

As we are given that 70.5 wt % aqueous perchloric acid that means 70.5 grams of perchloric acid present in 100 grams of solution.

Now we have to determine the mass of perchloric acid in 37.6 grams of aqueous perchloric acid.

As, 100 grams of aqueous perchloric acid (solution) contains 70.5 grams of perchloric acid.

So, 37.6 grams of aqueous perchloric acid (solution) contains \frac{37.6}{100}\times 70.5=26.5 grams of perchloric acid.

Thus, the mass of perchloric acid is, 26.5 grams.

Now we have to determine the mass of water are in the same solution.

Total mass of solution = 37.6 g

Mass of perchloric acid = 26.5 g

Mass of water = Total mass of solution - Mass of perchloric acid

Mass of water = 37.6 g - 26.5 g

Mass of water = 11.1 g

Thus, the mass of water in the same solution is, 11.1 grams

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3 years ago
The annual storage in Broad River watershed is 0 cm/y. Annual precipitation is 100 cm/y and evapotranspiration is 50 cm/y. The s
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0.34232

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Which option explains why Tyler is impressed in the following scenario?
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Which system provides an easier way for people to communicate with a computer than a graphical user interface (GUI)?
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C.

Explanation:

Natural language processing or NLP can be defined as an artificial intelligence that aids computers to understand, interpret, and manipulate human language. NLP is subfiled of many disciplines, for instance, artifical intelligence, linguistics, computer science, etc. NLP helps computers to interact with humans in their own language.

So, the easier way through which people can communicate with computers apart from GUI is NLP. Thus option C is correct.

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2 years ago
A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
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